I need to offer a 12V output to illuminate a 20mA LED, until a 12V input triggers (something) to make this output 20mA LED turn off. When the 12V trigger input is gone, the LED should turn back on.

I have designed a successful circuit to achieve this, but only using incandescent bulbs in my cars tail lamps, side markers, Headlamp, etc. I used the ground coming through the incandescent car bulbs to triggering (energizing) the BD140 transistors. When the 12v was applied from activating the car's indicators. The BD140 is deactivated, turning off the LED. I have attach the incandescent circuit to this post.

You will also note/see that I have a relay to accept the 0v/Ground trigger from the car's computer for when the 'Check Engine' light is active. I use the N.C. terminals on a relay to power the LED output, until the 0v/Ground trigger from the computer is activated. Once the relay is activated the 12v supplying the LED is removed, turning off the LED when the Check Engine is activated by the car's computer.

I have now replaced ALL my vehicles exterior lamps with LEDs. These new LEDs work fine. BUT ..... now my circuit does NOT. By replacing the incandescent with LEDs, I have removed the Ground triggering the BD140. ... The Check Engine light circuit still works. But I DO NOT want to hear relay clicking, so I do NOT want to use relays to solve this problem.

I tried to come up with a solution ... Circuit 2 (attached to this post) works fine for the Check Engine 0v/Ground trigger circuit. But Circuit 3 does not work well. The output in Circuit 3 causes the LED to light up very dim/faded. By lowering the R6 value from 1k to 100ohm, LED6 is very bright. But R6 at 100 ohms heats up within seconds. Once the BD139 is activated.

I am NOT opposed to using MOSfets or Darlington Transistors. I just have many BD139, BD140, 2n3904 and 2n3906.

I need to offer a 12V output to illuminate a 20mA LED, until a 12V input triggers (something) to make this output 20mA LED turn off. When the 12V trigger input is gone, the LED should turn back on.

I have designed a successful circuit to achieve this, but only using incandescent bulbs in my cars tail lamps, side markers, Headlamp, etc. I used the ground coming through the incandescent car bulbs to triggering (energizing) the BD140 transistors. When the 12v was applied from activating the car's indicators. The BD140 is deactivated, turning off the LED. I have attach the incandescent circuit to this post.

You will also note/see that I have a relay to accept the 0v/Ground trigger from the car's computer for when the 'Check Engine' light is active. I use the N.C. terminals on a relay to power the LED output, until the 0v/Ground trigger from the computer is activated. Once the relay is activated the 12v supplying the LED is removed, turning off the LED when the Check Engine is activated by the car's computer.

I have now replaced ALL my vehicles exterior lamps with LEDs. These new LEDs work fine. BUT ..... now my circuit does NOT. By replacing the incandescent with LEDs, I have removed the Ground triggering the BD140. ... The Check Engine light circuit still works. But I DO NOT want to hear relay clicking, so I do NOT want to use relays to solve this problem.

I tried to come up with a solution ... Circuit 2 (attached to this post) works fine for the Check Engine 0v/Ground trigger circuit. But Circuit 3 does not work well. The output in Circuit 3 causes the LED to light up very dim/faded. By lowering the R6 value from 1k to 100ohm, LED6 is very bright. But R6 at 100 ohms heats up within seconds. Once the BD139 is activated.

I am NOT opposed to using MOSfets or Darlington Transistors. I just have many BD139, BD140, 2n3904 and 2n3906.

For starters, you're not driving the LEDs correctly. You have 4 LEDs connected in parallel. The correct way to connect them is to have each one have it's own current limiting resistor. That will prevent current hogging which will likely result in a cascading failure destroying all of the LEDs.

Assuming 2V for the LED forward voltage, the dissipation in the resistor is P = V*V/R = 10V*10V/100 ohms = 1W. My guess is that you're using a 1/4W resistor...

If you replace the single 100 ohm resistor with 4 510 ohm resistors, each connected to one of the LEDs; you'll get about 20mA in each LED and each resistor will dissipate around 0.2W. You could use 1/4W resistors, but good design practice calls for derating them; so you should use 1/2W.

Oh yes. The 100 ohm Resistor in Circuit 3 gets very hot. Burns up actually. So I use a 1k ohm resistor in place of the 100 ohm. The Output LED is very dim, but the 1k resistor does NOT burn up.

TONY

Click to expand...

Of course the bloody resistor is going to get hot and possibly even burn up. Let's assume that the forward voltage drop across the LEDs is 2.5 volts. Your LEDs are rated at 20 mA so a 100 ohm resistor will give you the desired effect from a brightness standpoint, BUT you have to dissipate 9.5 volts x .095 amps of power (0.9 watts). You don't tell us the power dissipation rating of your resistors, but if you are using 1/4 watt or 1/2 watt resistors, you can bet your snake that they will burn up.
Off the top of my head there are a couple of avenues that you can pursue:
1. Use a resistor with a 5 watt rating and a heat sink. It'll get warm but wont burn up.
2. Use a 555 to pulse the LEDs at a 50% duty cycle. You will still probably need at least a 1 watt current limiting resistor.
Cheers

In your circuit, whatever the leds are on or off, the R6 always drawing a lots of current, and the leds in parallel directly was not good for leds themselves.

You can try the circuit below, I haven't try it yet, I was used the led drawing 80% of 20mA and that is 16mA, I also assuming that the led is 3V, R2, R4 just using 1/4W and R1, R3, R5~R8 using 1/2W or more.

In your new circuit, when the LED6(4 leds) is working then the current of R6 will be as :
I_R6 = (12V-3V)/(1K+100Ω)
= 9V/1.1K
= 8.1mA
If the led is 3V/20mA then the current can't be drive the led too light, but it is safe for leds.

When the LED6(4 leds) is not working then the current of R6 will be as :
I_R6 = V_R6*R6
= (12V-0.2V)/100Ω
= 11.8V/100Ω
= 118 mA, the current is waste too much.

Watts of R6:
W_R6 = V_R6*I_R6
= 11.8V*118mA
= 1.39 Watts

1.39W * 3 = 4.17 Watts
At least, you have to using 5 W resistor, but over 7 Watts is better for the bjt long time turn on.

When I designed the circuit, I forgot to calculate the affect of voltage divider, about the R2(10K) of the circuit in #8, in the beginning that I just want to use it to prevent something, but it will affecting the voltage, so I moved it to the Vbe as Rbe(R4) or you can take it away.

When I designed the circuit, I forgot to calculate the affect of voltage divider, about the R2(10K) of the circuit in #8, in the beginning that I just want to use it to prevent something, but it will affecting the voltage, so I moved it to the Vbe as Rbe(R4) or you can take it away.

Click to expand...

In my SIM your post #13 circuit works with and without R4. I will try on breadboard tonight. Thank you.

Is it possible the 27 ohm will get hot when the negative trigger is sent from cars computer. It will be a transistor negative output coming from the computer. That's why I had a bd139 and bd140 in the circuit. I was hoping to protect the computers transistor output by using a transistor as a buffer. Does my thought makes sense?

@stillgrowingup.
Sorry, I made a mistake, I forgot to calculate the current for your computer side, whatever reason that the resistor can't just connected to the computer like that, even it is a relay, because the current was too big, and I always be care about the current and heat, last night (my time zone) was too tired, when I woke up this morning and it really shocked me, I have to redesign the circuit.

You want to protect the computer side, that is correct, do you know what exactly the switch is(contacts of relay, bjt, mosfet)?

@ScottWang
I do NOT know what is driving output from Car computer. It is GM car from 1986. The output directly drives incandescent bulb and can remain lit/illuminated for hours at a time. I hope this info narrows the possibilities.

When I designed the circuit, I forgot to calculate the affect of voltage divider, about the R2(10K) of the circuit in #8, in the beginning that I just want to use it to prevent something, but it will affecting the voltage, so I moved it to the Vbe as Rbe(R4) or you can take it away.

Click to expand...

I have tested this circuit on my breadboard. The Output LED did NOT toggle until I increased R2 resistance from 1.2k to 15k. I did NOT try R2=10k ... Output led was nice and bright. ... When circuit was activated to shut OFF output LED. Output LED remain Dim/faded lit. R4=4.7k MUST be used to completely shut off output LED. Also, no components seemed to get warm/hot while testing. ... Should I use this circuit from post #13 with changes/modifications I have just stated?